Set-theoretic types and compile-time types

In this particular case I would probably do it like this:

defmodule TestModule do
  if Mix.env() == :prod do
    def prod?(), do: true
  else
    def prod?(), do: false
  end
end

There still seems to be a problem when this function is being used with “boolean” operators, for example:

defmodule Types do
  def prod?, do: unquote(Mix.env() == :prod)

  def a_or_b do
    if Types.prod?() and :rand.uniform() > 0.5 do
      :a
    else
      :b
    end
  end
end

warns

    warning: the following conditional expression will always evaluate to dynamic(false):

        Types.prod?()

    typing violation found at:
    │
  9 │     if Types.prod?() and :rand.uniform() > 0.5 do
    │     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    │
    └─ lib/types.ex:9: Types.a_or_b/0

Rewriting the caller like this seems to work but …

  def a_or_b do
    if Types.prod?() do
      if :rand.uniform() > 0.5 do
        :a
      end
    end || :b
  end

We will fix this on Elixir v1.18.1 We will make it so it only warns when comparing across different types.

@josevalim was this fix implemented?
Looks like there is still an issue with compile env in 1.18.1 (see screenshot below)

CleanShot 2024-12-30 at 12.46.422032×1367 129 KB

We use it more like this in our app to decide whether to start certain children in the supervision tree:

CleanShot 2024-12-30 at 12.55.022032×1367 159 KB

The fix right now only works when comparing constants. Because you encapsulate it in a function, it no longer works, but it would work if you did @env == :dev.

welp:

  defmacro always(term) do
    quote do
      :erlang.phash2(1, 1) == 0 && unquote(term)
    end
  end

  def prod?, do: always(unquote(Mix.env() == :prod))